04-BS-7 · May 2018

Question 1 of 13: Multi-Fluid Manometer on a Water Pipe

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

04-BS-7 Mechanics of Fluids — National Examination, 2018-May. Three (3) hours duration, closed book. Section A (Calculative, 9 questions, do 7) and Section B (Graphical & Analytical, 4 questions, do 3); every question is answered below regardless of the exam's "do N of M" instruction, so the set is a complete study resource.

Reference texts: White, F.M., Fluid Mechanics (8th ed.) — fluid statics and manometry (Ch. 2), Bernoulli and the energy equation (Ch. 3), viscous flow in ducts and the Moody chart (Ch. 6), flow past immersed bodies and drag (Ch. 7), potential flow and the Magnus effect (Ch. 8), open-channel flow and the hydraulic jump (Ch. 10), turbomachinery and jet propulsion (Ch. 11).

Check — assumptions used across this paper:

Q1's manometer chain is read off the extraction as a two-stage water–mercury–glycerine–(air)–glycerine–mercury system. The enclosed air pocket between the two glycerine columns is treated as weightless (uniform pressure), so only the one described open end is needed to close the hydrostatic chain back to pipe P; the second "opening" is not load-bearing for this calculation.
Q5's wave/hydraulic-jump analysis takes the depth "in front of the wave" (0.15 m, undisturbed, at rest) as the upstream state and "behind the wave" (0.75 m) as the downstream state, per the question's own prose (the raw figure-label ordering in the extraction is a reconstruction and is not used to override the stated text). The classic hydraulic-jump head-loss formula is applied to the given depths, and the swept flow rate uses the measured wave celerity directly — a standard engineering estimate, not a fully momentum-self-consistent bore solution.
Q6's air properties are taken at 20°C (domestic ambient, ρ=1.19 kg/m³) since no duct-air temperature is stated.
Q7(c)'s "discharged at right angles to the initial direction (20° becomes 0°)" is read as: the reverser's exhaust jet is normally angled 20° forward of the fully-radial (right-angle) direction; part (c) removes that forward lean entirely, leaving a purely radial (90° to the engine axis) discharge with zero axial velocity component.
Q8's terminal velocity and Q9's cable drag coefficient are obtained from the Reynolds-number relations the attached charts themselves plot (Morrison's sphere-drag correlation for Q8; the flat subcritical Cd≈1.2 plateau of the smooth-cylinder curve for Q9, since Re≈3×10⁴ falls solidly within it).

Question 1: Multi-Fluid Manometer on a Water Pipe (5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Pipe P above the water/mercury interface	100 mm
Mercury level, both arms of 1st U-tube	180 mm above trough
Glycerine top (air interface), 1st U-tube	360 mm above trough
Air-column depth to the bend	100 mm
Glycerine column, 2nd U-tube left arm	323 mm
Open mercury surface, 2nd U-tube right arm	30 mm below the 2nd tube's glycerine/Hg interface
Equivalent atmospheric head (open ends)	10 m of water
Specific gravities (Constants, p.11)	mercury 13.56, glycerine 1.26

Elevation ladder of the manometer chain (mm above the first U-tube's mercury trough): pipe P at 280, first Hg surfaces at 180, glycerine top/air interface at 360, air pocket to the bend at 460, second glycerine column down to 137, mercury down to the open surface at 107.

Find. The absolute pressure $P$ in pipe P (kPa).

Approach. Fix an elevation datum at the first U-tube's mercury trough, place every interface on it from the readings, then walk the hydrostatic law $p_{\text{down}}=p_{\text{up}}+\rho g\,\Delta z$ from the one clearly-open end (10 m of water atmospheric head) back to pipe P, switching density at each interface and treating the enclosed air pocket as weightless.

Elevations from the readings (datum = trough, z = 0). Water/Hg (1st, left) z=180 mm; pipe P z=280 mm; Hg/glycerine (1st, right) z=180 mm; glycerine/air (1st) z=360 mm; bend z=460 mm; air/glycerine (2nd) z=460 mm; glycerine/Hg (2nd) z=460−323=137 mm; open Hg surface z=137−30=107 mm.
Open end: equivalent atmospheric pressure. $$P_{\text{atm,eq}}=\rho_w g(10)=(1000)(9.81)(10)=98{,}100\ \text{Pa}$$
Up through mercury, 2nd tube (z=107→137). $$P_{137}=P_{\text{atm,eq}}-\rho_{Hg}g(0.137-0.107)=98{,}100-(13{,}560)(9.81)(0.030)=94{,}109\ \text{Pa}$$
Up through glycerine, 2nd tube (z=137→460). This equals the enclosed air pressure (weightless gas): $$P_{air}=94{,}109-(1260)(9.81)(0.323)=90{,}117\ \text{Pa}$$
Down through glycerine, 1st tube (z=360→180), starting from $P_{air}$ at z=360. $$P_{180}=90{,}117+(1260)(9.81)(0.180)=92{,}342\ \text{Pa}$$ Same elevation, same connected mercury body ⇒ this equals the pressure at the water/Hg interface (1st tube, left arm).
Up through water to pipe P (z=180→280). $$P=92{,}342-(1000)(9.81)(0.100)=\boxed{91{,}361\ \text{Pa}\approx 91.4\ \text{kPa}}$$

Quantity	Value
Enclosed air pressure	90.1 kPa
Pressure, 1st Hg/glycerine interface	92.3 kPa
Absolute pressure in pipe P	91.4 kPa

Concept & Theory

Topic. Multi-fluid, multi-U-tube manometry.

Key relations. Hydrostatic law $p_2=p_1+\rho g(z_1-z_2)$ applied leg-by-leg through a chain of immiscible fluids; equal pressure at equal elevation within one connected fluid body; negligible gas weight over small heights.

Why this works. Because the fluids are at rest, pressure depends only on elevation within each connected body, not on the tube's path — the whole manometer can be "unrolled" into a single elevation ladder and walked one interface at a time.

Common pitfall. Forgetting that a gas pocket carries the same pressure at both its interfaces (its own weight is negligible over the few hundred millimetres shown here), or mixing up which arm's reading applies to which fluid.

Source. White, Fluid Mechanics, Ch. 2 (Manometry).

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